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Determine the molarity of each of the solutions given below:
CBSE | Chemistry | Class 12 | Guides | NCERT | Solutions
Determine the molarity of each of the solutions given below:

(a) 30 g of Co(NO)3. 6H2O in 4.3 L of solution

(b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.

We are aware of this.,

Molarity =\frac{\text { Moles of Solute }}{\text { Volume of solution in litre }}

(a) Molar mass of \mathrm{Co}(\mathrm{NO})_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}=59+2(14+3 \times 16)+6 \times 18=291 \mathrm{~g} \mathrm{~mol}^{-1}

so, Moles of \mathrm{Co}(\mathrm{NO})_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}=\frac{30}{291} \mathrm{~mol}

=0.103 \mathrm{~mol}

so, molarity =\frac{0.103 \mathrm{~mol}}{4.3 L}

=0.023 \mathrm{M}

(b) Total number of moles in 1000 \mathrm{~mL} of 0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}=0.5 \mathrm{~mol}

So, Number of moles found in 30 \mathrm{~mL} of 0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}=\frac{0.5 \times 30}{1000} \mathrm{~mol}

=0.015 \mathrm{~mol}

Hence, molarity =\frac{0.015}{0.5 L} mol =0.03 \mathrm{M}

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